Intuitive definitions of the Orbit and the Stabilizer

The orbit of $x$ is "everything that can be reached from $x$ by an action of something in $G$."

The stabilizer of $x$ is "the set of all elements of $G$ which don't move $x$ when they act on $x$".

Those already seem pretty intuitive... what else can be said?

I guess you might want to look at orbits and stabilizers for particular actions. For example, if a group is acting on itself by conjugation, then the orbit of an element is that element's conjugacy class. One element stabilizes another in this action exactly when they commute.

Some examples may illuminate these definitions.

Let $G$ be the circle group, $$G = \{z\in\mathbb{C}: |z|=1\}.$$ This is the unit circle in $\mathbb{C}$, and it is a group under multiplication. By the same token, $G$ acts on $X = \mathbb{C}$ by multiplication. Geometrically, multiplication by $z = e^{i\theta}$ acts on $\mathbb{C}$ by rotation by the angle $\theta$. Now if we fix some $x\in\mathbb{C}$, the orbit through $x$ will be exactly the circle of radius $|x|$ centred at the origin -- unless $x=0$, in which case the orbit will just be the point $\{0\}$.

How about stabilizers? Let $x\in\mathbb{C}$, and let $G^x$ be its stabilizer. If $x\neq 0$, then $zx = x$ if and only if $z=1$, and so $G^x=\{1\}$. On the other hand, if $x=0$ then $zx = x$ for all $z\in G$, and so $G^x=G$.

Here's an example with more interesting stabilizers. Let $G$ now be the dihedral group of order 8,

$$G = \{\rho,\tau: \rho^4=e, \tau^2=e, \tau\rho\tau = \rho^{-1}\}.$$

This acts on the set $X = \{1,2,3,4\}$ of vertices of a square

with $\rho$ acting by rotation counterclockwise by $\frac \pi 4$ radians and $\tau$ acting by reflection through the red line. In this way, the elements of $G$ consist of four rotations (through $0, \frac\pi 4, \frac\pi 2$, and $\frac{3\pi} 4$ radians) and four reflections (horizontally, vertically and through each of the two diagonals). Note that each vertex can be sent to any other given vertex by an appropriate rotation, and so the orbit of each vertex is $\{1,2,3,4\}$. What is the stabilizer of, say, $3$? None of the nontrivial rotations fix $3$, and the only reflection that fixes $3$ is $\tau$, and so the stabilizer of $3$ is $\{e,\tau\}$.

A bit about group actions:

In algebra and geometry, a group action is a description of symmetries of objects using groups. The essential elements of the object are described by a set, and the symmetries of the object are described by the symmetry group of this set, which consists of bijective transformations of the set. In this case, the group is also called a permutation group (especially if the set is finite or not a vector space) or transformation group (especially if the set is a vector space and the group acts like linear transformations of the set).

A group action is an extension to the definition of a symmetry group in which every element of the group "acts" like a bijective transformation (or "symmetry") of some set, without being identified with that transformation. This allows for a more comprehensive description of the symmetries of an object, such as a polyhedron, by allowing the same group to act on several different sets of features, such as the set of vertices, the set of edges and the set of faces of the polyhedron.

If $G$ is a group and $X$ is a set then a group action may be defined as a group homomorphism $h$ from $G$ to the symmetric group of $X$. The action assigns a permutation of $X$ to each element of the group in such a way that the permutation of X assigned to the identity element of $G$ is the identity (do-nothing) transformation of $X$; a product gh of two elements of $G$ is the composition of the permutations assigned to $g$ and $h$.

Since each element of $G$ is represented as a permutation, a group action is also known as a permutation representation.

See also Gowers's blog entry for a "down to earth" discussion on group actions

• Group Actions (i)

Oribts

The defining properties of a group guarantee that the set of orbits of (points x in) $X$ under the action of $G$ form a partition of X. The associated equivalence relation is defined by saying $x \sim y$ if and only if there exists a $g \in G$ with $gx = y.$ The orbits are then the equivalence classes under this relation; two elements $x$ and $y$ are equivalent if and only if their orbits are the same; i.e., $Gx = Gy$.

If you have the spare time, you might find this You Tube video, Orbits of group action helpful.

Fixed points and stabilizer subgroups

Given $g \in G$ and $x \in X$ with $gx=x$, we say $x$ is a fixed point of $g$ and $g$ fixes $x$.

For every $x \in X$, we define the stabilizer subgroup of $x$ as the set of all elements in $G$ that fix $x$:

$$G_x = \{g\in G\mid gx = x\}.$$

$G_x$ is a subgroup of $G$, though typically not a normal one.

Also, Wikipedia discusses orbits and stabilizers and how they relate, in its "Group Action" entry.

See also Gowers's follow-up blog entry for a "down to earth" discussion on group actions, orbits, and stabilizers:

• Group Actions ii: orbit-stabilizer-theorem

Orbits

Consider a sphere $S \subset \mathbb{R^3}$ and a group $G$ of (all) rotations along the OZ axis (north-south pole, as Earth).

$\hspace{70pt}$

For every angle $\alpha \in [0, 2\pi)$ there is an element of the group $g_\alpha$ that would map a point $x$ of the sphere to the point $y = g_\alpha\cdot x$ such that the directed angle between would be $\alpha$. Observe that $S$ is a 3d object, where the group is almost like $[0,2\pi)$ - those are two different things, yet are joined by the "$\cdot$" operator, i.e. the group $G$ acts on sphere $S$.

What is an orbit then? See for yourself: $Orb(x) = \{ gx \mid g \in G\}$. That means the orbit of a point $x$ of the sphere $S$ is a set of all the points that a $G$ (i.e. rotations) can produce from it, in our case a circle, like the dotted one on the picture. Can you guess why it is called an orbit?

Stabilizers

Stabilizers are subgroups that won't disturb your points. In our sphere example, we have two kind of points: poles and non-poles. Does any rotation moves a pole? No, so $G_{\text{pole}} = G$. However, every other point would be disturbed by any non-zero rotation, hence the stabilizer of a non-pole point is a trivial group, $G_{\text{non-pole}} = \{1\}$.

In a more involved example, group of permutations of $\{A,B,C\}$, we can pick some element, let it be $B$, and still have a non-trivial stabilizer $G_{B} = \{(A)(B)(C), (AC)(B)\} \sim \mathbb{Z}_2$.

In conclusion: stabilizer of $x$ is the biggest subgroup that won't disturb your $x$.

I hope it explained something ;-)